On convergence/divergence of improper integral with hyperbolic function

Question

I am trying to determine whether $\int_0^{\infty}{(\frac{1}{xsinh(x)}-\frac{1}{x})dx}$ converges or diverges. It seems like inevitably divergent in 0 point. But how to show it? Maybe it should be compared with something? Will be grateful for your ideas.

Answer 1

Note that there are two concerns; the behavior of the integrand at the lower and upper limits. And either or both could render the integral divergent.

Let's examine the behavior of the integrand near $x=0$. Note that we have

$$\sinh(x)=x+O(x^3)$$

and therefore

$$\frac{1}{x\sinh(x)}-\frac1x=\frac1x\left(\frac{1}{x+O(x^3)}-1\right)=\frac1{x^2}-\frac1x +O(1)$$

Inasmuch as $\int_0^1 \frac{1}{x^\alpha}\,dx$ diverges for $\alpha \ge 1$, the integral of interest diverges.

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Note that if the lower limit were $a>0$ instead of $0$, the behavior of the integrand as $x\to \infty$ is heuristically

$$\frac{1}{x\sinh(x)}-\frac{1}{x}\sim \frac{2e^{-x}-1}{x}\sim -\frac1x $$

and since the integral $\int_a^\infty \frac1x \,dx$ diverges, so does the integral $\int_a^\infty \left(\frac{1}{x\sinh(x)}-\frac1x\right)\,dx$